📚 Formulae Derivation from OxfordAQA 9630 PH05 WRE Jun23 | OxfordAQA 9630 PH05 WRE Jun23 公式推导
This article presents step-by-step derivations of key formulae that appear in the OxfordAQA 9630 PH05 June 2023 paper, using the Worked Example Resource (WRE) as a reference. Understanding these derivations is essential for mastering the concepts of circular motion, simple harmonic motion, capacitance, electromagnetic induction, and gravitational fields.
本文以 OxfordAQA 9630 PH05 2023年6月试卷的工作示例资源（WRE）为参考，逐步推导关键公式。掌握这些推导过程对于理解圆周运动、简谐运动、电容、电磁感应和引力场等概念至关重要。
1. Derivation of Centripetal Acceleration | 向心加速度的推导
Consider an object moving at constant speed v along a circular path of radius r. In a small time Δt, the object moves from point P to Q, sweeping out an angle Δθ. The velocity vectors at P and Q have equal magnitude but different directions; the change in velocity is Δv.
考虑一个物体以恒定速率 v 沿半径为 r 的圆形路径运动。在很短的 Δt 时间内，物体从 P 点运动到 Q 点，扫过的角度为 Δθ。P 点和 Q 点的速度矢量大小相等但方向不同，速度的变化量为 Δv。
These two velocity vectors form an isosceles triangle with sides of length v. The displacement vector from P to Q has magnitude Δs = r Δθ. By vector triangle similarity, the ratio of the magnitude of Δv to v is equal to the ratio of Δs to r, so |Δv| / v = Δs / r. Substituting Δs gives |Δv| = (v / r) × r Δθ = v Δθ.
这两个速度矢量构成一个腰长为 v 的等腰三角形。从 P 到 Q 的位移矢量大小为 Δs = r Δθ。通过矢量三角形的相似性，Δv 的大小与 v 之比等于 Δs 与 r 之比，即 |Δv| / v = Δs / r。代入 Δs 得到 |Δv| = (v / r) × r Δθ = v Δθ。
|Δv| / v = Δs / r → |Δv| = v Δθ
Dividing by Δt and taking the limit as Δt → 0 gives the instantaneous acceleration magnitude: a = v (dθ/dt) = v ω, where ω = dθ/dt is the angular speed. Since v = r ω, we obtain the centripetal acceleration in two standard forms:
将上式除以 Δt 并取 Δt → 0 时的极限，得到瞬时加速度大小：a = v (dθ/dt) = v ω，其中 ω = dθ/dt 是角速率。利用 v = r ω，便得到向心加速度的两个标准形式：
a = v² / r
a = r ω²
The direction of this acceleration is always towards the centre of the circle, hence the term ‘centripetal’.
该加速度的方向始终指向圆心，因此被称为“向心”加速度。
2. Relationship Between Angular Speed and Period | 角速率与周期的关系
For uniform circular motion, one complete revolution corresponds to an angular displacement of 2π radians. If the period is T, the angular speed ω is defined as the rate of change of angular displacement.
对于匀速圆周运动，完整转一圈对应的角位移为 2π 弧度。如果周期为 T，则角速率 ω 定义为角位移的变化率。
ω = 2π / T
Since the object travels the circumference 2π r in time T with constant speed v, we also have v = 2π r / T. Combining the two relations confirms v = r ω.
由于物体在 T 时间内以恒定速率 v 走过周长 2π r，因此也有 v = 2π r / T。将两式结合即可验证 v = r ω。
3. Deriving the Displacement Equation in Simple Harmonic Motion | 简谐运动中位移方程的推导
Simple harmonic motion (SHM) is the projection of uniform circular motion onto a diameter. Consider a point P moving around a circle of radius A (the amplitude) with constant angular speed ω. Let the phase angle at t = 0 be φ. The horizontal displacement x of the projection onto the x-axis is given by the x-coordinate of P.
简谐运动（SHM）是匀速圆周运动在某一直径上的投影。考虑一个点 P 以恒定的角速率 ω 沿半径为 A（振幅）的圆周运动，设 t = 0 时的相位角为 φ。投影到 x 轴上的水平位移 x 即为 P 的 x 坐标。
x = A cos(θ) where θ = ω t + φ
Thus the displacement–time equation is:
因此位移–时间方程为：
x = A cos(ω t + φ)
If the oscillation starts with maximum positive displacement, φ = 0 and x = A cos(ω t). Alternatively, using a sine function for a different starting condition, x = A sin(ω t + φ’) where φ’ accounts for the phase shift. The most general form is x = A sin(ω t + φ₀) ±, depending on convention.
若振荡从最大正位移开始，则 φ = 0，x = A cos(ω t)。也可以在初始条件不同时使用正弦函数，x = A sin(ω t + φ’)，其中 φ’ 反映了相位差。最通用的形式为 x = A sin(ω t + φ₀) 或 x = A cos(ω t + φ₀)，视习惯而定。
4. Deriving Velocity and Acceleration in SHM | 简谐运动速度与加速度的推导
The velocity v of the SHM is the time derivative of displacement. Using x = A cos(ω t + φ), we differentiate to obtain:
简谐运动的速度 v 是位移对时间的导数。利用 x = A cos(ω t + φ)，求导得到：
v = dx/dt = -A ω sin(ω t + φ)
The acceleration a is the derivative of velocity:
加速度 a 是速度的导数：
a = dv/dt = -A ω² cos(ω t + φ) = -ω² x
Therefore, the defining equation of SHM is a ∝ -x. The negative sign indicates that acceleration is always directed towards the equilibrium position. A useful relation between speed and displacement can be found by eliminating time using the identity sin²(θ) + cos²(θ) = 1:
因此，简谐运动的定义方程为 a ∝ -x。负号表明加速度始终指向平衡位置。通过恒等式 sin²(θ) + cos²(θ) = 1 消去时间，可得到速率与位移之间的有用关系：
v = ± ω √(A² – x²)
The maximum speed occurs at x = 0: v_max = ω A, and the maximum acceleration occurs at x = ± A: a_max = ω² A.
最大速率出现在 x = 0 处：v_max = ω A；最大加速度出现在 x = ± A 处：a_max = ω² A。
5. Energy in a Simple Harmonic Oscillator | 简谐振子的能量推导
For a mass–spring system, the restoring force is F = -k x, where k is the spring constant. The potential energy stored is the work done against this force to stretch the spring from 0 to x:
对于质量-弹簧系统，回复力为 F = -k x，其中 k 为劲度系数。储存的势能等于克服该力将弹簧从 0 拉伸到 x 所做的功：
E_p = ∫₀ˣ k x dx = ½ k x²
Since the motion is SHM, we have ω = √(k / m), so k = m ω². Thus E_p = ½ m ω² x². The kinetic energy is E_k = ½ m v² = ½ m ω² (A² – x²) using v² = ω² (A² – x²).
因为运动是 SHM，所以 ω = √(k / m)，即 k = m ω²。因此 E_p = ½ m ω² x²。动能 E_k = ½ m v² = ½ m ω² (A² – x²)（利用了 v² = ω² (A² – x²)）。
The total mechanical energy E_total is constant and equals the sum of kinetic and potential energies:
总机械能 E_total 为常数，且等于动能与势能之和：
E_total = E_k + E_p = ½ m ω² (A² – x²) + ½ m ω² x² = ½ m ω² A²
This shows that the total energy depends only on the amplitude and the angular frequency.
这表明总能量仅取决于振幅和角频率。
6. Deriving the Capacitor Discharge Equation | 电容器放电方程的推导
A capacitor C charged to an initial voltage V₀ discharges through a resistor R. At any instant, the capacitor voltage V and the resistor voltage I R must sum to zero around the loop: V – I R = 0 (taking the direction of discharge current appropriately). Since V = Q / C and I = -dQ/dt (because charge Q decreases as the capacitor discharges), we obtain:
一个充电至初始电压 V₀ 的电容器 C 通过电阻 R 放电。任意时刻，回路中电容器电压 V 与电阻电压 I R 之和为零：V – I R = 0（适当选取放电电流的方向）。因为 V = Q / C 且 I = -dQ/dt（放电时电荷 Q 在减少），我们得到：
Q / C + R (dQ/dt) = 0 → dQ/dt = -Q / (R C)
This is a first-order differential equation. Separating variables and integrating gives:
这是一个一阶微分方程。变量分离并积分得到：
∫ (1/Q) dQ = -∫ 1/(R C) dt
ln Q = -t / (R C) + constant
At t = 0, Q = Q₀ (initially stored charge), so the constant is ln Q₀. Therefore,
在 t = 0 时，Q = Q₀（初始储存电荷），所以常数为 ln Q₀。因此，
ln (Q / Q₀) = -t / (R C) → Q = Q₀ e⁻ ᵗ⁄ᴿᶜ
Using V = Q / C, the voltage decays as V = V₀ e⁻ ᵗ⁄ᴿᶜ. The time constant τ = R C is the time for the charge (or voltage) to fall to 1/e ≈ 37% of its initial value.
运用 V = Q / C，电压按 V = V₀ e⁻ ᵗ⁄ᴿᶜ 衰减。时间常数 τ = R C 是电荷（或电压）降至初始值 1/e ≈ 37% 所需的时间。
7. Deriving the Radius of Curvature for a Charged Particle in a Magnetic Field | 磁场中带电粒子圆周运动半径的推导
When a particle of charge q moves with velocity v perpendicular to a uniform magnetic field of flux density B, it experiences a magnetic force F_B = B q v (from Lorentz force law, F = q v B sin θ, with θ = 90°). This force is always perpendicular to the velocity, providing the centripetal force required for circular motion.
当电荷量为 q 的粒子以速度 v 垂直于磁感应强度为 B 的匀强磁场运动时，它受到磁力 F_B = B q v（来自洛伦兹力公式 F = q v B sin θ，且 θ = 90°）。该力始终与速度垂直，提供圆周运动所需的向心力。
B q v = m v² / r
Rearranging for the radius r of the circular path gives:
整理可得圆周路径的半径 r：
r = m v / (B q)
The period T of the orbit is the time to complete one revolution at speed v: T = 2π r / v. Substituting r yields T = 2π m / (B q), which is independent of speed – a key result for applications such as the cyclotron.
轨道周期 T 是以速度 v 转动一圈的时间：T = 2π r / v。代入 r 得到 T = 2π m / (B q)，该周期与速率无关——这是回旋加速器等应用的关键结论。
8. Faraday’s Law of Electromagnetic Induction – Deriving the Induced EMF | 法拉第电磁感应定律——感应电动势的推导
Faraday’s law states that the magnitude of the induced emf in a circuit equals the rate of change of magnetic flux linkage. For a coil of N turns, the flux linkage Φ_link = N Φ, where Φ = B A cos θ is the magnetic flux through one turn.
法拉第定律指出，回路中感应电动势的大小等于磁通量链环的变化率。对于 N 匝线圈，磁通量链环 Φ_link = N Φ，其中 Φ = B A cos θ 是穿过单匝的磁通量。
If the coil is rotated with angular speed ω in a uniform field B, the angle between the area vector and B changes as θ = ω t. Thus Φ = B A cos(ω t). The flux linkage is N B A cos(ω t). Differentiating with respect to time gives the induced emf:
如果线圈在匀强磁场 B 中以角速率 ω 旋转，面积矢量与 B 之间的夹角按 θ = ω t 变化。因此 Φ = B A cos(ω t)。磁通量链环为 N B A cos(ω t)。对时间求导即得感应电动势：
ε = – d(N Φ) / dt = – N B A d[cos(ω t)] / dt = N B A ω sin(ω t)
The negative sign (Lenz’s law) indicates that the induced emf opposes the change in flux. The peak emf ε₀ = N B A ω is often used in the analysis of AC generators.
负号（楞次定律）表明感应电动势阻碍磁通量的变化。峰值电动势 ε₀ = N B A ω 常用于交流发电机的分析。
9. Gravitational Potential and Field Strength | 引力势与引力场强度的推导
The gravitational field strength g at a distance r from a point mass M (or outside a spherical mass) is defined as the force per unit mass on a small test mass m: g = F/m. From Newton’s law of gravitation, F = G M m / r², so
距离点质量 M（或球对称质量外部）r 处的引力场强度 g 定义为作用在小试验质量 m 上的单位质量受力：g = F / m。由牛顿万有引力定律 F = G M m / r²，可得
g = G M / r²
Gravitational potential V_g is the work done per unit mass to bring a test mass from infinity to that point without changing its kinetic energy. The work done against the gravitational force F = G M m / r² over a displacement dr (towards the mass) is dW = -F dr = – (G M m / r²) dr. The potential is this work per unit mass:
引力势 V_g 是使单位质量从无穷远移动到该点且不改变其动能所做的功。移动质量 m 时反抗引力 F = G M m / r²，沿 dr（向质量方向）所做功为 dW = -F dr = – (G M m / r²) dr。引力势即此功的单位质量值：
V_g = W/m = ∫_∞^r – (G M / r²) dr = [G M / r]_∞^r = – G M / r
The negative sign indicates that the potential is taken as zero at infinity and becomes more negative as we approach the mass. The relationship between field and potential is g = – dV_g/dr, which can be verified by differentiating -G M / r with respect to r.
负号表明将无穷远处的势取为零，当我们接近质量时，势变得更负。场与势的关系为 g = – dV_g / dr，这可以通过对 -G M / r 求导来验证。
10. Deriving the Period of a Simple Pendulum | 单摆周期的推导
A simple pendulum of length L, displaced by a small angle θ, experiences a restoring force along the arc of magnitude m g sin θ. For small angles, sin θ ≈ θ (in radians), so the tangential acceleration a_t = -g θ. Noting that the arc displacement s = L θ, we get a_t = d²s/dt². The equation of motion becomes:
一个长度为 L 的单摆，偏离一个小角度 θ，沿着圆弧受到大小为 m g sin θ 的回复力。对于小角度，sin θ ≈ θ（弧度制），所以切向加速度 a_t = -g θ。注意弧位移 s = L θ，有 a_t = d²s/dt²。运动方程变为：
d²s/dt² = – (g / L) s
This is analogous to a = -ω² x in SHM, with ω² = g / L. Hence
这类似于简谐运动中的 a = -ω² x，其中 ω² = g / L。因此
ω = √(g / L)
Since the period T = 2π / ω, we obtain the familiar result:
因为周期 T = 2π / ω，我们得到熟悉的结果：
T = 2π √(L / g)
The derivation highlights that for small oscillations the period is independent of amplitude (isochronous) and depends only on pendulum length and gravitational field strength.
该推导突显了小角度振荡时周期与振幅无关（等时性），仅取决于摆长和重力场强度。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导